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It would simply be the irrational square root of a rational number. There is no special name for it.

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Q: If x is irrational what would x be if x squared is rational?

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No. If x is irrational, then x/x = 1 is rational.

Yes - if I had an irrational number x, and I added that to the number (7-x), I would end up with 7.If the number is irrational, it can be subtracted from a rational/integer to make another irrational.

Let q be a non-zero rational and x be an irrational number.Suppose q*x = p where p is rational. Then x = p/q. Then, since the set of rational numbers is closed under division (by non-zero numbers), p/q is rational. But that means that x is rational, which contradicts x being irrational. Therefore the supposition that q*x is rational must be false ie the product of a non-zero rational and an irrational cannot be rational.

Yes.

It is irrational.

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If x is rational the it is rational. If x is irrational then it is irrational.

Suppose a is rational (and non-zero) and x is irrational. Suppose ax is rational;write ax = b where b is rational.Then x = b/a, and x would be rational, contradiction.

I don't knw all the possible numbers, but π is an example. π2 is irrational such as π.

No. If x is irrational, then x/x = 1 is rational.

Yes - if I had an irrational number x, and I added that to the number (7-x), I would end up with 7.If the number is irrational, it can be subtracted from a rational/integer to make another irrational.

Let q be a non-zero rational and x be an irrational number.Suppose q*x = p where p is rational. Then x = p/q. Then, since the set of rational numbers is closed under division (by non-zero numbers), p/q is rational. But that means that x is rational, which contradicts x being irrational. Therefore the supposition that q*x is rational must be false ie the product of a non-zero rational and an irrational cannot be rational.

Yes.

Suppose x is a rational number and y is an irrational number.Let x + y = z, and assume that z is a rational number.The set of rational number is a group.This implies that since x is rational, -x is rational [invertibility].Then, since z and -x are rational, z - x must be rational [closure].But z - x = y which implies that y is rational.That contradicts the fact that y is an irrational number. The contradiction implies that the assumption [that z is rational] is incorrect.Thus, the sum of a rational number x and an irrational number y cannot be rational.

an irrational number PROOF : Let x be any rational number and y be any irrational number. let us assume that their sum is rational which is ( z ) x + y = z if x is a rational number then ( -x ) will also be a rational number. Therefore, x + y + (-x) = a rational number this implies that y is also rational BUT HERE IS THE CONTRADICTION as we assumed y an irrational number. Hence, our assumption is wrong. This states that x + y is not rational. HENCE PROVEDit will always be irrational.

Let R1 = rational number Let X = irrational number Assume R1 + X = (some rational number) We add -R1 to both sides, and we get: -R1 + x = (some irrational number) + (-R1), thus X = (SIR) + (-R1), which implies that X, an irrational number, is the sum of two rational numbers, which is a contradiction. Thus, the sum of a rational number and an irrational number is always irrational. (Proof by contradiction)

The polynomial is not factorable with rational numbers. If you want to use irrational numbers it would be x minus the square root of 2 times x plus the square root of 2.

Yes, because when x equals 1, the square root of x is rational and the square root of -x is irrational, and when x equals -1, the square root of x is irrational and the square root of -x is rational.

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